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Date	May Example question	Marks available	13	Reference code	EXM.2.AHL.TZ0.28
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 0
Command term	Use, State, and Calculate	Question number	28	Adapted from	N/A

Question

The random variable X is thought to follow a binomial distribution B (4, $p$ ). In order to investigate this belief, a random sample of 100 observations on X was taken with the following results.

An automatic machine is used to fill bottles of water. The amount delivered, $Y$ ml, may be assumed to be normally distributed with mean $\mu$ ml and standard deviation 8 ml. Initially, the machine is adjusted so that the value of $\mu$ is 500. In order to check that the value of $\mu$ remains equal to 500, a random sample of 10 bottles is selected at regular intervals, and the mean amount of water, ${\bar y}$ , in these bottles is calculated. The following hypotheses are set up.

H₀: $\mu$ = 500; H₁: $\mu$ ≠ 500

The critical region is defined to be $\left( {\bar y < 495} \right) \cup \left( {\bar y > 505} \right)$ .

State suitable hypotheses for testing this belief.

[1]

a.i.

Calculate the mean of these data and hence estimate the value of $p$ .

[5]

a.ii.

Calculate an appropriate value of ${\chi ^2}$ and state your conclusion, using a 1% significance level.

[13]

a.iii.

Find the significance level of this procedure.

[5]

b.i.

Some time later, the actual value of $\mu$ is 503. Find the probability of a Type II error.

[3]

b.ii.

Markscheme

H₀: The data are B (4, $p$ ); H₁ : The data are not B (4, $p$ ) A1

[1 mark]

a.i.

Mean $= \frac{{1 \times 32 + \ldots + 4 \times 14}}{{100}}$ M1A1

= 1.8 A1

$4\widehat p = 1.8 \Rightarrow \widehat p = 0.45$ M1A1

[5 marks]

a.ii.

The expected frequencies are

A1A1A1A1A1

The last two classes must be combined because the expected frequency for $x$ = 4 is less than 5. R1

${\chi ^2} = \frac{{{{17}^2}}}{{9.15}} + \frac{{{{32}^2}}}{{29.95}} + \frac{{{{19}^2}}}{{36.75}} + \frac{{{{32}^2}}}{{24.15}} - 100$ M2

= 18.0 A2

DF = 2 (A1)

Critical value = 9.21 A1

We conclude, at the 1% significance level, that X does not fit a binomial model. R1

Special case: award the following marks to candidates who do not combine classes.

${\chi ^2} = \frac{{{{17}^2}}}{{9.15}} + \frac{{{{32}^2}}}{{29.95}} + \frac{{{{19}^2}}}{{36.75}} + \frac{{{{18}^2}}}{{20.05}} + \frac{{{{14}^2}}}{{4.1}} - 100$ M2

= 39.6 A0

DF = 3 (A1)

Critical value = 11.345 A1

We conclude, at the 1% significance level, that X does not fit a binomial model. R1

[13 marks]

a.iii.

Under H₀, the distribution of ${\bar y}$ is N (500, 6.4). (A1)

Significance level = P ${\bar y}$ < 495 or > 505 | H₀ M2

= 2 × 0.02405 (A1)

= 0.0481 A1 N5

[5 marks]

b.i.

The distribution of ${\bar y}$ is now N (503, 6.4). (A1)

P(Type ΙΙ error) = P(495 < ${\bar y}$ < 505) (M1)

= 0.785 A1 N3

[3 marks]

b.ii.

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

a.iii.

[N/A]

b.i.

[N/A]

b.ii.

Syllabus sections

Topic 4—Statistics and probability » SL 4.8—Binomial distribution

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